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Posers and Puzzles

Posers and Puzzles

  1. 31 Jan '08 10:54
    I found out that 1^5-1 is dividable with 5. So is 2^5-2 and 3^5-3.
    Is 2008^5-2008 also dividable with 5?
    Is perhaps all n^5-n dividable with 5? Show me this is the case, or not.
  2. Standard member adam warlock
    Baby Gauss
    31 Jan '08 12:27
    Originally posted by FabianFnas
    I found out that 1^5-1 is dividable with 5. So is 2^5-2 and 3^5-3.
    Is 2008^5-2008 also dividable with 5?
    Is perhaps all n^5-n dividable with 5? Show me this is the case, or not.
    I experimented with it a bit and it seems likely that you got yourself a theorem. With all numbers I tried for n the result was always the form of a number ending in 0 and numbers that end in 0 are divisible by 5. But I can't get way to write your expresion in a product that has 10 in it.

    This is all I got so far
    n^5-n=n*(n^4-1)=n*(n^2+1)*(n^2-1)=n*(n^2+1)*(n+1)*(n-1)
    Now n*(n+1) is the result of an arithmetic summation and I can't see if that help us or not but nonetheless it is interesting. And (n^2+1)*(n-1) maybe is something to though I don't know what...

    I'll try for a bit longer and tell you about updates.
  3. Standard member wolfgang59
    Infidel
    31 Jan '08 13:12
    Originally posted by FabianFnas
    I found out that 1^5-1 is dividable with 5. So is 2^5-2 and 3^5-3.
    Is 2008^5-2008 also dividable with 5?
    Is perhaps all n^5-n dividable with 5? Show me this is the case, or not.
    Is 2008^5-2008 also dividable with 5? YES

    2008^5 ends with the digit 8.
    (since considering last digits 8*8 = 64, 4*8 = 32, 2*8 = 16, 6*8=48)

    Subtracting 2008 gives us a number ending in zero and hence divisible by 5.

    There are only 10 cases to consider; numbers ending in 0,1,2, ..9. Prove for each of these and your theorem is proven.
  4. Standard member adam warlock
    Baby Gauss
    31 Jan '08 13:14
    Originally posted by wolfgang59
    There are only 10 cases to consider; numbers ending in 0,1,2, ..9. Prove for each of these and your theorem is proven.
    I like your divide and conquer strategy
  5. 31 Jan '08 15:24
    It's clearly true if the number you start with is divisible by 5.
    Fermat says a^4 = 1 (mod 5) if a not divisible by 5.
    So a^5 = a (mod 5), whence the result.
  6. Standard member Agerg
    The 'edit'or
    31 Jan '08 15:46 / 1 edit
    Proof by induction that for all n >= 0, 5|(n^5 - n)

    Base case
    let n = 0
    0^5 - 0 = 0 is divisible by 5 so result holds

    Assume by inductive hypothesis result holds for n = k, ie; 5|(k^5-k)

    Let n = k+1

    (K+1)^5 - (k+1)
    =k^5 +5k^4 +10k^3 + 10k^2 + 5k +1 - (k+1)
    =k^5 +5(k^4 +2k^3 +2k^2 + k) - k
    =(k^5 - k) + 5(k^4 +2k^3 +2k^2 + k) is divisible by 5 (inductive hypothesis)
    Hence result holds for n = k+1

    Therefore, by induction on n it is true that for all n >= 0, 5|(n^5 - n)

    a similar argument can be used for n < 0
  7. 31 Jan '08 15:47 / 1 edit
    Originally posted by Tricky Dicky
    It's clearly true if the number you start with is divisible by 5.
    Fermat says a^4 = 1 (mod 5) if a not divisible by 5.
    So a^5 = a (mod 5), whence the result.
    If a not divisable by 5, hmm, what about when a is divisable by 5? Like 12345^5-12345 ?

    But even if you don't know Fermat and what he says you can come to the correct result quite easy.

    Edit: ... and now I see that Agerg has came up with a correct result. Well done!
  8. 31 Jan '08 17:36 / 1 edit
    Originally posted by FabianFnas
    If a not divisable by 5, hmm, what about when a is divisable by 5? Like 12345^5-12345 ?

    But even if you don't know Fermat and what he says you can come to the correct result quite easy.

    Edit: ... and now I see that Agerg has came up with a correct result. Well done!
    well, clearly when a is divisible by 5, a=0 (mod 5) and so a^5=0 (mod 5)... so a^5-a = 0 (mod 5). The nonzero modulo classes are interesting because we can invoke Fermat's Little Theorem, a^p-1 = 1 (mod p) for any prime p. I think this is what Tricky Dicky was saying.

    In fact, this proves that it is true for ALL prime number exponents, not just 5. Since a^(p-1)=1 (mod p), we see that a^p = a (mod p) and a^p - a = 0 (mod p).

    So 12345^71-12345 is divisible by 71... and so forth. Fermat's "Little Theorem" is quite powerful

    Edit: note, similarly to what Tricky Dicky said, a^p-1 = 1 (mod p) is only true when p does not divide a. but if it does, a^p = 0 (mod p) and a = 0 (mod p)... so the result a^p - a = 0 (mod p) still holds
  9. Subscriber coquette
    Already mated
    31 Jan '08 18:00
    I'm so not a mathematician but I think that this problem is simple and obvious. the number raised to any power with the number subtracted is obviously still a multiple of the number. no fancy formulas or anything else is needed, nor is it a theorem, at least any more than close the window there's a draft in here is a theorem.
  10. Standard member adam warlock
    Baby Gauss
    31 Jan '08 18:38
    Originally posted by coquette
    I'm so not a mathematician but I think that this problem is simple and obvious. the number raised to any power with the number subtracted is obviously still a multiple of the number. no fancy formulas or anything else is needed, nor is it a theorem, at least any more than close the window there's a draft in here is a theorem.
    n^p-n is still a multiple of n like you say but it is not obvious at all, at least not to me that n^p-n is a multiple of p.

    And sometimes on math the most difficult things to prove are the obvious ones. If you know calculus try to prove the intermediat value for yourself. Pretty self-evident and prety obvious. At first sight it even looks like not being a theorem at all. But just try to prove it for yourself.
  11. Subscriber coquette
    Already mated
    31 Jan '08 19:09
    I'm not a mathemetician, so I have to do this like i'm just talking.

    Take any number "n"

    Raise to some multiple of n - ANY multiple of n.

    subtract one of the Ns that you raised it to, it's still a multiple of n.

    That's too easy.
  12. 31 Jan '08 19:15
    Originally posted by coquette
    I'm so not a mathematician but I think that this problem is simple and obvious. the number raised to any power with the number subtracted is obviously still a multiple of the number. no fancy formulas or anything else is needed, nor is it a theorem, at least any more than close the window there's a draft in here is a theorem.
    you're correct that this is obvious... what ISN'T obvious is that it's a multiple of the POWER you raised it to. that's the interesting fact: 3^71 - 3 is a multiple of 71 (in addition to the trivial fact it is a multiple of 3)
  13. 31 Jan '08 21:14
    Originally posted by Aetherael
    you're correct that this is obvious... what ISN'T obvious is that it's a multiple of the POWER you raised it to. that's the interesting fact: 3^71 - 3 is a multiple of 71 (in addition to the trivial fact it is a multiple of 3)
    What does that ^ symbol mean?
  14. Standard member wolfgang59
    Infidel
    31 Jan '08 22:07
    to the power of

    hence 10^2 = 10 x 10 = 100
  15. Standard member Agerg
    The 'edit'or
    01 Feb '08 00:19 / 3 edits
    Originally posted by coquette
    I'm so not a mathematician but I think that this problem is simple and obvious. the number raised to any power with the number subtracted is obviously still a multiple of the number. no fancy formulas or anything else is needed, nor is it a theorem, at least any more than close the window there's a draft in here is a theorem.
    but we are looking for divisibility by the index...not the number raised to the index
    for example, consider 2^4-2...this is not divisible by 4 yet 2^5 -2 *is* divisible by 5...Yes m^n-m is certainly divisible by m but that is of no interest here....is it divisible by n??!
    That is what we are interested in!